# The Physics Philes, lesson 99: I Have a Need. A Need for Wave Speed!

Over the past several weeks we’ve been learning about mechanical waves in all they’re wavy wonder. Today we’ll look a little more in depth at one key property of a wave: the wave speed. We’ll start off easy with the propagation of a transverse wave on a string. Don’t worry though; these concepts will translate to other kinds of mechanical waves.

There are two physical quantities that determine the wave speed of a transverse wave on a string: the tension of the string and its linear mass density (otherwise known as mass per unit length). Increasing the tension should increase the restoring force that straighten the string, thus increasing the wave speed. If we increase the mass, the motion should become more sluggish and the wave speed should decrease. We can show this is true using two different methods. Today we’ll look at the first, simpler method. Tomorrow we’ll take a look at the other, more general method.

OK, let’s think about a perfectly flexible string. In equilibrium, the tension is F and the linear mass density is μ. Since we’re all physicists here, we can ignore the weight of the string. In this case, when the string is in equilibrium we an assume that it makes a perfectly straight line. It might look something like this: At time 0, imagine that you apply a constant force upward on the left end of the string. Like this: This upward force causes more and more mass to move. Notice point P. That’s the division point between the moving and non-moving portions. The wave is traveling at a constant speed, so the division point moves with that constant speed.

All of the particles in the motion portion are moving at a constant velocity, but not a constant acceleration. Why? Let’s think about the impulse of F_y. The impulse of F_y at time t is the product of the force’s y-component and the time. We know that, according to the impulse-momentum theorem, that the impulse is equal to the change in the total transverse component of momentum of the moving part. Since the system started with no transverse momentum, the total momentum at any specific time is This tells us that the total momentum must increase proportionally with time. But also notice that, since the point P moves with a constant speed, the total mass in motion is also proportionally change with time. That means that the change in momentum is associated with the amount of mass in motion, not any increasing velocity.

OK, now we’re ready to derive the wave speed equation. To do this, we’re going to use similar triangles and the impulse-momentum theorem. Here is a drawing of just the moving part of diagram above: Notice that it’s a right triangle. We’ve established that the transverse impulse is equal tot he change in transverse momentum. The impulse of the transverse force in any time is the product of the two. Now it’s time to use that triangle. You can see that the black triangle with sides vt and v_yt is similar to the triangle with sides F_y and F. So the relationship we have here is  The transverse impulse ends up being The mass of the moving string is the product of its linear mass density and length. Transverse momentum is the product of the mass and the vertical velocity, so the equation we come up with is Even though the momentum change is due to a change in the amount of mass that is moving and not a change in the mass’s velocity, the impulse is still equal to the total change of momentum of the system. We can use that relationship to our advantage. We get Now all that’s left is to solve for v. Huh. Look at that. The vertical speed doesn’t appear in this equation, so we can determine that the wave speed doesn’t depend on that. This equation was derived from a special case, but it can be applied to any transverse wave on a string. Cool, right?

OK, so that was the easy method of finding the wave speed. Not too bad, right? Just some algebra. Next week it’s going to get a bit trickier, and, frankly, I’m a little afraid. But I have confidence that we can do it. So bone up on your partial derivatives and I’ll see you back here next week.

Featured image credit: Jeff Rowley via Flickr